# MATH 409 Lecture 25

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Lecture Slides

Course Evaluations.

Final exam this Friday, 15:00–17:00

Extra Office Hours:

• 4 Dec 12:30–14:00
• 5 Dec 12:00–14:00
• 6 Dec 10:00–12:00

# Final Review

## Topic List

### Part 1: Axiomatic Model of the Real Numbers

Chapters 1.1–1.6, Appendix A

### Part 2: Limits and Continuity

Chapters 2.1–2.5, 3.1–3.4

Chapters 4.1–4.5

Chapters 5.1–5.4

Chapters 6.1–6.4

## Theorems to Know

### Archimedian Principle

For any real number ${\displaystyle \epsilon >0}$, there exists a ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle n\,\epsilon >1}$.

Theorem. ${\displaystyle \mathbb {Z} }$, ${\displaystyle \mathbb {Q} }$, ${\displaystyle \mathbb {N} \times \mathbb {N} }$ are countable

Theorem. ${\displaystyle \mathbb {R} }$ is uncountable.

### Theorems on Limits

Squeeze Theorem. If ${\displaystyle \lim _{n\to \infty }x_{n}=\lim _{n\to \infty }y_{n}=a}$, and ${\displaystyle x_{n} for all ${\displaystyle n>N}$, then ${\displaystyle \lim _{n\to \infty }w_{n}=a}$.

Theorem. Any monotone sequence converges to a limit if bounded, and diverges to infinity otherwise.

Theorem. Any Cauchy sequence is convergent.

### Theorems on Derivatives

Theorem. If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable at ${\displaystyle a\in \mathbb {R} }$, then their sum ${\displaystyle f+g}$, difference ${\displaystyle f-g}$, and product ${\displaystyle f\cdot g}$ are also differentiable at ${\displaystyle a}$. Moreover,

{\displaystyle {\begin{aligned}(f+g)'(a)&=f'(a)+g'(a)\\(f-g)'(a)&=f'(a)-g'(a)\\(f\cdot g)'(a)&=f'(a)\,g(a)+f(a)\,g'(a)\\\end{aligned}}}

If, additionally, ${\displaystyle g(a)\neq 0}$, then the quotient ${\displaystyle {\frac {f}{g}}}$ is also differentiable at ${\displaystyle a}$ and

${\displaystyle \left({\frac {f}{g}}\right)'(a)={\frac {f'(a)\,g(a)-f(a)\,g'(a)}{(g(a))^{2}}}}$

Mean Value Theorem. If a function ${\displaystyle f}$ is continuous on ${\displaystyle \left[a,b\right]}$ and differentiable on ${\displaystyle \left(a,b\right)}$, then there exists ${\displaystyle c\in (a,b)}$ such that ${\displaystyle f(b)-f(a)=f'(c)\,(b-a)}$.

### Theorems on Integrals

Theorem. [Linearity]. If ${\displaystyle f}$ and ${\displaystyle g}$ are integrable on ${\displaystyle \left[a,b\right]}$, then ${\displaystyle f+g}$ is also integrable on ${\displaystyle \left[a,b\right]}$ and

${\displaystyle \int _{a}^{b}\left(f(x)+g(x)\right)\,\mathrm {d} x=\int _{a}^{b}f(x)\,\mathrm {d} x+\int _{a}^{b}g(x)\,\mathrm {d} x}$

Theorem. If a function ${\displaystyle f}$ is integrable on ${\displaystyle [a,b]}$, then for any ${\displaystyle c\in (a,b)}$,

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x=\int _{a}^{c}f(x)\,\mathrm {d} x+\int _{c}^{b}f(x)\,\mathrm {d} x}$

### Theorems on Series

Integral Test. Suppose that ${\displaystyle f:\left[1,\infty \right)\to \mathbb {R} }$ is positive and decreasing on ${\displaystyle \left[1,\infty \right)}$. Then the series ${\displaystyle \sum _{n=1}^{\infty }f(n)}$ converges if and only if the function ${\displaystyle f}$ is improperly integrable on ${\displaystyle \left[1,\infty \right)}$.

Ratio Test. Let ${\displaystyle \left\{a_{n}\right\}}$ be a sequence of reals with ${\displaystyle a_{n}\neq 0}$ for large ${\displaystyle n}$. Suppose that a limit ${\displaystyle r=\lim _{n\to \infty }{\frac {\left|a_{n+1}\right|}{\left|a_{n}\right|}}}$ exists (or is finite), then

1. If ${\displaystyle r<1}$, then ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges absolutely.
2. If ${\displaystyle r>1}$, then ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ diverges.

## Sample Problems

### Problem 1 (20 pts)

Union of countable sets is also countable

Theorem. Suppose ${\displaystyle E_{1}}$, ${\displaystyle E_{2}}$, ${\displaystyle E_{3}}$, ... are countable sets. Prove that their union ${\displaystyle E_{1}\cup E_{2}\cup E_{3}\cup \dots }$ is also a countable set.

Proof. First we are going to show that ${\displaystyle \mathbb {N} \times \mathbb {N} }$ is countable.

Consider a relation ${\displaystyle \prec }$ on the set ${\displaystyle \mathbb {N} \times \mathbb {N} }$ such that ${\displaystyle (n_{1},n_{2})\prec (m_{1},m_{2})}$ if and only if either ${\displaystyle n_{1}+n_{2} or else ${\displaystyle n_{1}+n_{2}=m_{1}+m_{2}}$ and ${\displaystyle n_{1}. (similar to a lexicographic order, but important difference). It is easy to see that ${\displaystyle \prec }$ is a strict linear order. [1] Moreover, for any pair ${\displaystyle (m_{1},m_{2})\in \mathbb {N} \times \mathbb {N} }$, there are only finitely many pairs ${\displaystyle (n_{1},n_{2})}$ such that ${\displaystyle (n_{1},n_{2})\prec (m_{1},m_{2})}$. It follows that ${\displaystyle \prec }$ is a well-ordering. [2]

Now we define inductively a mapping ${\displaystyle F:\mathbb {N} \to \mathbb {N} \times \mathbb {N} }$ such that for any ${\displaystyle n\in \mathbb {N} }$, the pair ${\displaystyle F(n)}$ is the least (relative to ${\displaystyle \prec }$) pair different from ${\displaystyle F(k)}$ for all natural numbers ${\displaystyle k. It follows from the construction that ${\displaystyle F}$ is bijective. The inverse mapping ${\displaystyle F^{-1}}$ can be given explicitly by

${\displaystyle F^{-1}(n_{1},n_{2})={\frac {1}{2}}\,\left(n_{1}+n_{2}-2\right)\,\left(n_{1}+n_{2}-1\right)+n_{1}}$

Thus ${\displaystyle \mathbb {N} \times \mathbb {N} }$ is a countable set.

Now suppose ${\displaystyle E_{1},E_{2},\ldots }$ are countable sets. Then for any ${\displaystyle n\in \mathbb {N} }$, there exists a bijective mapping ${\displaystyle f_{n}:\mathbb {N} \to E_{n}}$. Let us define a map ${\displaystyle g:\mathbb {N} \times \mathbb {N} \to E_{1}\cup E_{2}\cup \dots }$ by ${\displaystyle g(n_{1},n_{2})=f_{n_{1}}(n_{2})}$. Obviously ${\displaystyle g}$ is onto.

Since the set ${\displaystyle \mathbb {N} \times \mathbb {N} }$ is countable, there exists a sequence ${\displaystyle p_{1},p_{2},p_{3},\ldots }$ that forms a complete list of its elements. Then the sequence ${\displaystyle g(p_{1}),g(p_{2}),g(p_{3}),\ldots }$ contains all elements of the union ${\displaystyle E_{1}\cup E_{2}\cup E_{3}\cup \dots }$. Although the latter sequence may include repetitions, we can choose a subsequence ${\displaystyle \left\{g(p_{n_{k}})\right\}}$ in which every element of the union appears exactly once. Note that the subsequence is infinite since each of the sets ${\displaystyle E_{1},E_{2},\ldots }$ is infinite.

Now the map ${\displaystyle h:\mathbb {N} \to E_{1}\cup E_{2}\cup E_{3}\cup \dots }$ defined by ${\displaystyle h(k)=g(p_{n_{k}})}$ for ${\displaystyle k=1,2,\ldots }$ is a bijection.

quod erat demonstrandum

### Problem 2 (20 pts)

Evaluate following limits

#### Subproblem 2a

${\displaystyle \lim _{x\to 0}\log {\left({\frac {1}{1+\cot {x^{2}}}}\right)}}$

This function can be represented as the composition of 4 functions:

1. ${\displaystyle f_{1}(x)=x^{2}}$
2. ${\displaystyle f_{2}(y)=\cot {y}}$
3. ${\displaystyle f_{3}(z)=(1+z)^{-1}}$
4. ${\displaystyle f_{4}(u)=\log {u}}$

Since ${\displaystyle f_{1}}$ is continuous, we have ${\displaystyle \lim _{x\to 0}f_{1}(x)=f_{1}(0)=0}$. Moreover, ${\displaystyle f_{1}(x)>0}$ for ${\displaystyle x\neq 0}$.

Since ${\displaystyle \lim _{y\to 0+}f_{2}(x)=+\infty }$, it follows that ${\displaystyle (f_{2}\circ f_{1})(x)\to +\infty }$ as ${\displaystyle x\to 0}$.

Further, ${\displaystyle f_{3}(z)\to 0+}$ as ${\displaystyle z\to +\infty }$ and ${\displaystyle f_{4}(u)\to -\infty }$ as ${\displaystyle u\to 0^{+}}$.

Finally ${\displaystyle (f_{4}\circ f_{3}\circ f_{2}\circ f_{1})(x)\to -\infty }$ as ${\displaystyle x\to 0}$.

#### Subproblem 2b

${\displaystyle \lim _{x\to 64}{\frac {{\sqrt {x}}-8}{{\sqrt[{3}]{x}}-4}}}$

This is indeterminate of form ${\displaystyle {\frac {0}{0}}}$, but we can do better: Consider function ${\displaystyle u(x)=x^{\frac {1}{6}}}$ defined on ${\displaystyle \left(0,\infty \right)}$. Since this function is continuous at 64 and ${\displaystyle u(64)=2}$, we obtain

{\displaystyle {\begin{aligned}\lim _{x\to 64}{\frac {{\sqrt {x}}-8}{{\sqrt[{3}]{x}}-4}}&=\lim _{x\to 64}{\frac {\left(u(x)\right)^{3}-8}{\left(u(x)\right)^{2}-4}}\\&=\lim _{y\to 2}{\frac {y^{3}-8}{y^{2}-4}}\\&=\lim _{y\to 2}{\frac {(y-2)(y^{2}+2y+4)}{(y-2)(y+2)}}\\&=\lim _{y\to 2}{\frac {y^{2}+2y+4}{y+2}}\\&=3\end{aligned}}}

#### Subproblem 2c

${\displaystyle \lim _{n\to \infty }\left(1+{\frac {c}{n}}\right)^{n}}$, where ${\displaystyle c\in \mathbb {R} }$.

Let ${\displaystyle a_{n}=\left(1+{\frac {c}{n}}\right)^{n}}$ for ${\displaystyle n\in \mathbb {N} }$.

For ${\displaystyle n}$ large enough, we have ${\displaystyle 1+{\frac {c}{n}}>0}$, so ${\displaystyle a_{n}>0}$. Then

${\displaystyle \log {\left(\left(1+{\frac {c}{n}}\right)^{n}\right)}=n\log \left(1+{\frac {c}{n}}\right)=\left.{\frac {\log {\left(1+c\,x\right)}}{x}}\right|_{x={\frac {1}{n}}}}$

Since ${\displaystyle {\frac {1}{n}}\to 0}$ as ${\displaystyle n\to \infty }$ and

{\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {\log {\left(1+c\,x\right)}}{x}}&=\left.\left(\log {\left(1+c\,x\right)}\right)'\right|_{x=0}\\&=\left.{\frac {c}{1+c\,x}}\right|_{x=0}\\&=c\end{aligned}}}

We obtain that ${\displaystyle \log _{a_{n}}\to c}$ as ${\displaystyle n\to \infty }$, therefore ${\displaystyle a_{n}=\mathrm {e} ^{\log {a_{n}}}=\mathrm {e} ^{c}}$ as ${\displaystyle n\to \infty }$

### Problem 3 (20 pts)

Prove that series converges to ${\displaystyle \sin {x}}$ for any ${\displaystyle x\in \mathbb {R} }$:

${\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}\,{\frac {x^{2n-1}}{(2n-1)!}}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}}$

Proof. The function ${\displaystyle f(x)=\sin {x}}$ is infinitely differentiable on ${\displaystyle \mathbb {R} }$. According to Taylor's formula, for any ${\displaystyle x,x_{0}\in \mathbb {R} }$ and ${\displaystyle n\in \mathbb {N} }$,

${\displaystyle f(x)=f(x_{0})+{\frac {f'(x_{0})}{1!}}\,(x-x_{0})+\dots +{\frac {f^{(n)}(x_{0})}{n!}}\,(x-x_{0})^{n}+R_{n}(x,x_{0})}$

where ${\displaystyle R_{n}(x,x_{0})={\frac {f^{(n+1)}(\theta )}{(n+1)!}}\,(x-x_{0})^{n+1}}$ for some ${\displaystyle \theta =\theta (x,x_{0})}$ between ${\displaystyle x}$ and ${\displaystyle x_{0}}$. Since ${\displaystyle f'(x)=\cos {x}}$ and ${\displaystyle f''(x)=-\sin {x}=-f(x)}$ for all ${\displaystyle x\in \mathbb {R} }$, it follows that ${\displaystyle f^{(n+1)}(\theta )\leq 1}$ for all ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle \theta \in \mathbb {R} }$. Further, one derives that ${\displaystyle R_{n}(x,x_{0})\to 0}$ as ${\displaystyle n\to \infty }$. Thus we obtain an expansion of ${\displaystyle \sin {x}}$ into a series.

In the case ${\displaystyle x_{0}=0}$, this is the required series. (up to zero terms)

quod erat demonstrandum

### Problem 4 (20 pts)

Evaluate following integrals

#### Subproblem 4a

${\displaystyle \int {\frac {\sqrt {1+{\sqrt[{4}]{x}}}}{2{\sqrt {x}}}}\,\mathrm {d} x}$

To find this integral, we change the variable twice:

1. ${\displaystyle \int {\frac {\sqrt {1+{\sqrt[{4}]{x}}}}{2{\sqrt {x}}}}\,\mathrm {d} x=\int {\sqrt {1+{\sqrt {u}}}}\,\mathrm {d} u}$ for ${\displaystyle u={\sqrt {x}}}$
2. ${\displaystyle \int {\sqrt {1+{\sqrt {u}}}}\,\mathrm {d} u=\int (4w^{4}-4w^{2})\,\mathrm {d} w}$ for ${\displaystyle w={\sqrt {1+{\sqrt {u}}}}}$

The integral of this is ${\displaystyle {\frac {4}{5}}w^{5}-{\frac {4}{3}}w^{3}+C={\frac {4}{3}}\,\left(1+x^{\frac {1}{4}}\right)^{\frac {3}{2}}+C}$.

#### Subproblem 4b

${\displaystyle \int _{0}^{\sqrt {3}}{\frac {x^{2}+6}{x^{2}+9}}\,\mathrm {d} x}$

To evaluate this definite integral, we use linearity of the integral and a substitution ${\displaystyle x=3u}$:

{\displaystyle {\begin{aligned}\int _{0}^{\sqrt {3}}{\frac {x^{2}+6}{x^{2}+9}}\,\mathrm {d} x&=\int _{0}^{\sqrt {3}}\left(1-{\frac {3}{x^{2}+9}}\right)\,\mathrm {d} x\\&=\int _{0}^{\sqrt {3}}1\,\mathrm {d} x-\int _{0}^{\sqrt {3}}{\frac {3}{x^{2}+9}}\,\mathrm {d} x\\&={\sqrt {3}}-\int _{0}^{\frac {\sqrt {3}}{3}}{\frac {3}{(3u)^{2}+9}}\,\mathrm {d} (3u)\\&={\sqrt {3}}-\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{u^{2}+1}}\,\mathrm {d} u\\&={\sqrt {3}}-\left.\arctan {u}\right|_{u=0}^{{\frac {1}{\sqrt {3}}}A}\\&={\sqrt {3}}-{\frac {\pi }{6}}-0\end{aligned}}}

#### Subproblem 4c

${\displaystyle \int _{0}^{\infty }x^{2}\,\mathrm {e} ^{-x}\,\mathrm {d} x}$

To evaluate this improper integral, we integrate by parts twice:

{\displaystyle {\begin{aligned}\int _{0}^{\infty }x^{2}\,\mathrm {e} ^{-x}\,\mathrm {d} x&=-\int _{0}^{\infty }x^{2}\,\mathrm {d} \left(\mathrm {e} ^{-x}\right)\\&=\left.-x^{2}\,\mathrm {e} ^{-x}\right|_{0}^{\infty }+\int _{0}^{\infty }\mathrm {e} ^{-x}\,\mathrm {d} (x^{2})\\&=\int _{0}^{\infty }2x\,\mathrm {e} ^{-x}\,\mathrm {d} x\\&=\left.-2x\,\mathrm {e} ^{-x}\right|_{0}^{\infty }+\int _{0}^{\infty }\mathrm {e} ^{-x}\,\mathrm {d} (2x)\\&=\int _{0}^{\infty }2\mathrm {e} ^{-x}\,\mathrm {d} x\\&=\left.-2\mathrm {e} ^{-x}\right|_{0}^{\infty }\\&=2\end{aligned}}}

### Problem 5 (20 pts)

Check for convergence of series

#### Subproblem 5a

${\displaystyle \sum _{n=1}^{\infty }{\frac {{\sqrt {n+1}}-{\sqrt {n}}}{{\sqrt {n+1}}+{\sqrt {n}}}}}$

This diverges because the sumand terms are just ${\displaystyle {\frac {1}{\left({\sqrt {n+1}}+{\sqrt {n}}\right)^{2}}}>{\frac {1}{4(n+1)}}}$. The series of the other term diverges because it is the harmonic series, therefore the main series diverges by comparison.

#### Subproblem 5b

${\displaystyle \sum _{n=1}^{\infty }{\frac {{\sqrt {n}}+2^{n}\,\cos {n}}{n!}}}$

This series can be represented as ${\displaystyle \sum _{n=1}^{\infty }\left(b_{n}+c_{n}\,\cos {n}\right)}$, where ${\displaystyle b_{n}={\frac {\sqrt {n}}{n!}}}$ and ${\displaystyle c_{n}={\frac {2^{n}}{n!}}}$ for all ${\displaystyle n\in \mathbb {N} }$.

The series ${\displaystyle \sum _{n=1}^{\infty }b_{n}}$ and ${\displaystyle \sum _{n=1}^{\infty }c_{n}}$ both converge due to ratio test, and so does their sum. ${\displaystyle \left|b_{n}+c_{n}\,\cos {n}\right|\leq b_{n}+c_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$, so the series converges absolutely by the comparison test.

### Subproblem 5c

${\displaystyle \sum _{n=2}^{\infty }{\frac {\left(-1\right)^{n}}{n\,\log {n}}}}$

This series converges by the alternating series test, but not absolutely due to the integral test.

### Bonus Problem 6 (15 pts)

Prove that an infinite product converges:

${\displaystyle \prod _{n=1}^{\infty }{\frac {n^{2}+1}{n^{2}}}}$

Take log of product:

${\displaystyle \log {\left(\prod _{n=1}^{\infty }{\frac {n^{2}+1}{n^{2}}}\right)}=\sum _{n=1}^{\infty }\log {\left({\frac {n^{2}+1}{n^{2}}}\right)}}$

## Footnotes

1. A strict linear order has transitivity. In this case, ${\displaystyle \prec }$ is also a total ordering.
2. A well-ordering implies that any finite set must have a minimum and maximum element. In our case, ${\displaystyle \mathbb {N} }$ is bounded below, so any set, finite or infinite, must have a minimum element.